Problem1 - CISL: Communication and Information Systems ...

Problem1

(a) false

(b) true

(c) true

(d) true

(e) true

(f) false

(g) false

(h) true

(i) true

( j) false

Problem2

(1) epsilon

(2) zeta

(3) eta

(4) lamda

(5) nu

(6) phi

(7) chi

(8) theta

(9) gamma

(10) sigma

(11) Lamda

(12) Pi

(13) phi

(14) Omega

Problem 3

(a)

(b)

(c)

(d)

(e)

Problem4

(a)

(b)

(c)

(d)

(e)

Problem 5. (8 points: p. 359, A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd Edition. Prentice Hall, 1996 ) In this problem, we derive the DTFT from the DTFS. Fill in the blanks.

Consider a general sequence x[n] that is of finite duration. That is, for some integers 1N and 2N , x[n]=0 outside

the range 1 2N n N− ≤ ≤ . From this aperiodic signal, we can construct a periodic sequence [ ]x n for which x[n] is

one period. As we choose the period N to be larger, [ ]x n is identical to x[n] over a longer interval, and as N →∞ , [ ] [ ]x n x n= for any finite value of n.

Let us now examine the Fourier series representation of [ ]x n , we have

(2 / )(1) [ ] ,jk N nk

k Nx n a e π

=< >

= ∑

(2 / )1(2) [ ] jk N nk

n Na x n e

Nπ−

=< >

= ∑ .

Since [ ] [ ]x n x n= over a period that includes the interval 1 2N n N− ≤ ≤ , it is convenient to choose the interval of

summation in eq. (2) to include this interval, so that [ ]x n can be replaced by [ ]x n in the summation. Therefore

2

1

(2 / ) (2 / )1 1(3) [ ] [ ] ,N

jk N n jk N nk

n N na x n e x n e

N Nπ π

∞− −

=− =−∞

= =∑ ∑

where in the second equality in eq. (3) we have used the fact that [ ]x n is zero outside the interval 1 2N n N− ≤ ≤ .

Defining the function

(4) ( ) [ ] ,j n

nX j x n e ωω

∞−

=−∞

= ∑

we see the coefficient ka are proportional to samples of ( )X jω , i.e.,

01(5) ( ),jk

ka X eN

ω=

where 0 2 / Nω π= is the spacing of the samples in the frequency domain. Combining eqs. (1) and (5) yields

0 01(6) [ ] ( ) .jk jk n

k Nx n X e e

Nω ω

=< >

= ∑

Since 0 2 / Nω π= , or equivalently, 01/ / 2N ω π= , eq. (6) can be rewritten as

0 00

1(7) [ ] ( ) .2

jk jk n

k Nx n X e eω ω ω

π =< >

= ∑

Each term in the summation in eq. (7) represents the area of a rectangle of height 0 0( )jk jk nX e eω ω and width 0ω . As

0 0ω → , the summation becomes an integral. Furthermore, since the summation is carried out over N consecutive

intervals of width 0 2 / Nω π= , the total interval of integration will always have a width of 2π . Therefore, as

N →∞ , [ ] [ ]x n x n= , and eq. (7) becomes

2

1(8) [ ] ( )2

j j nx n X e e dω ω

πω

π= ∫

where, since ( )j j nX e eω ω is periodic with period 2π , the interval of integration can be taken as any interval of length 2π . Thus, we have the following pair of equations:

2

1(9) [ ] ( )2

j j nx n X e e dω ω

πω

π= ∫ ,

(10) ( ) [ ] .j j n

nX e x n eω ω

∞−

=−∞

= ∑

Problem 7. (15 points: p. 355, A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd Edition. Prentice Hall, 1996 ) (a) Design a compensatory system that, when provided with the output of the measuring device, produces an output equal to the instantaneous temperature of the liquid. Sol) We consider a compensatory system that produces an output equal to instantaneous temperature. Since the step response is /2 /2( ) (1 ) ( ) (1 ) 0 ,t ts t e u t e for t− −= − = − ≤ < ∞

the impulse response has to be /2 /21 1( ) ( ) ( 0 ) ( ).2 2

t tdh t s t e for t e u tdt

− −= = ≤ < ∞ =

So, the frequency response of the system is

1/ 2( ) ( ) .1

2

FTh t H jj

ωω

→ =+

We now desire to build an inverse (compensatory) for the above system. Therefore, the frequency response of the inverse system (compensatory system) has to be

1 1( ) 2 .( ) 2

G j jH j

ω ωω

= = +

Taking the inverse Fourier transform we obtain 1( ) ( ) 2 ( )g t t u tδ= + where 1( )u t unit doublet (lecture 4, p8) is. (b) Consider a measuring device whose overall output can be modeled as the sum of the response of the measuring device characterized by eq. (1) and an interfering noise signal ( )n t . Such a model is depicted in Figure (a). Suppose that ( ) sin .n t tω= What is the contribution of ( )n t to the output of the inverse system, and how does this output changes as ω is increased? Sol) When ( ) sinn t tω= passes through the inverse system at problem (a), the output will be

( ) ( ) ( ) sin 2 cos .y t n t g t t tω ω ω= ∗ = +

We see that the output is directly proportional to ω . Therefore, as ω increase the contribution to the output due to the noise also increases. (c) Suppose that we wish to design a compensatory system that will slow down the response to actual temperature variations, but also will attenuate the noise ( )n t . Let the impulse response of this system be ( ) ( ).ath t ae u t−= Choose a so that the overall system of Figure (b) responds as quickly as possible to a step change in temperature, subject to the constraint that the amplitude of the portion of the output due to the noise ( ) sin 6n t t= is no larger than 1/4. Sol) We consider the output due to the noise ( ) sin 6 .n t t= So, the output will be

( ) ( ) ( ) ( ) ( ) ( ).FTy t n t h t Y j N j H jω ω ω= ∗ → = In this case we require the amplitude of the portion of the output,

( ) 1( )( ) 4

Y jH j

N jω

ωω

= ≤ when 6ω = .

Since 2

22 2( ) ,aH j

aω

ω=

+ we require that

2

2 2

1 .36 16

aa

≤+

Therefore, 6 .15

a ≤